3.149 \(\int \frac{x (a+b \tanh ^{-1}(c x))}{d+e x} \, dx\)

Optimal. Leaf size=156 \[ -\frac{b d \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right )}{2 e^2}+\frac{b d \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 e^2}+\frac{d \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e^2}-\frac{d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^2}+\frac{a x}{e}+\frac{b \log \left (1-c^2 x^2\right )}{2 c e}+\frac{b x \tanh ^{-1}(c x)}{e} \]

[Out]

(a*x)/e + (b*x*ArcTanh[c*x])/e + (d*(a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/e^2 - (d*(a + b*ArcTanh[c*x])*Log[(
2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/e^2 + (b*Log[1 - c^2*x^2])/(2*c*e) - (b*d*PolyLog[2, 1 - 2/(1 + c*x)])/
(2*e^2) + (b*d*PolyLog[2, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*e^2)

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Rubi [A]  time = 0.153831, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.412, Rules used = {5940, 5910, 260, 5920, 2402, 2315, 2447} \[ -\frac{b d \text{PolyLog}\left (2,1-\frac{2}{c x+1}\right )}{2 e^2}+\frac{b d \text{PolyLog}\left (2,1-\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{2 e^2}+\frac{d \log \left (\frac{2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e^2}-\frac{d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e^2}+\frac{a x}{e}+\frac{b \log \left (1-c^2 x^2\right )}{2 c e}+\frac{b x \tanh ^{-1}(c x)}{e} \]

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*ArcTanh[c*x]))/(d + e*x),x]

[Out]

(a*x)/e + (b*x*ArcTanh[c*x])/e + (d*(a + b*ArcTanh[c*x])*Log[2/(1 + c*x)])/e^2 - (d*(a + b*ArcTanh[c*x])*Log[(
2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/e^2 + (b*Log[1 - c^2*x^2])/(2*c*e) - (b*d*PolyLog[2, 1 - 2/(1 + c*x)])/
(2*e^2) + (b*d*PolyLog[2, 1 - (2*c*(d + e*x))/((c*d + e)*(1 + c*x))])/(2*e^2)

Rule 5940

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rule 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 5920

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTanh[c*x])*Log[2/(1
+ c*x)])/e, x] + (Dist[(b*c)/e, Int[Log[2/(1 + c*x)]/(1 - c^2*x^2), x], x] - Dist[(b*c)/e, Int[Log[(2*c*(d +
e*x))/((c*d + e)*(1 + c*x))]/(1 - c^2*x^2), x], x] + Simp[((a + b*ArcTanh[c*x])*Log[(2*c*(d + e*x))/((c*d + e)
*(1 + c*x))])/e, x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 - e^2, 0]

Rule 2402

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> -Dist[e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rule 2447

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[(Pq^m*(1 - u))/D[u, x]]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rubi steps

\begin{align*} \int \frac{x \left (a+b \tanh ^{-1}(c x)\right )}{d+e x} \, dx &=\int \left (\frac{a+b \tanh ^{-1}(c x)}{e}-\frac{d \left (a+b \tanh ^{-1}(c x)\right )}{e (d+e x)}\right ) \, dx\\ &=\frac{\int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{e}-\frac{d \int \frac{a+b \tanh ^{-1}(c x)}{d+e x} \, dx}{e}\\ &=\frac{a x}{e}+\frac{d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1+c x}\right )}{e^2}-\frac{d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}-\frac{(b c d) \int \frac{\log \left (\frac{2}{1+c x}\right )}{1-c^2 x^2} \, dx}{e^2}+\frac{(b c d) \int \frac{\log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{1-c^2 x^2} \, dx}{e^2}+\frac{b \int \tanh ^{-1}(c x) \, dx}{e}\\ &=\frac{a x}{e}+\frac{b x \tanh ^{-1}(c x)}{e}+\frac{d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1+c x}\right )}{e^2}-\frac{d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}+\frac{b d \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e^2}-\frac{(b d) \operatorname{Subst}\left (\int \frac{\log (2 x)}{1-2 x} \, dx,x,\frac{1}{1+c x}\right )}{e^2}-\frac{(b c) \int \frac{x}{1-c^2 x^2} \, dx}{e}\\ &=\frac{a x}{e}+\frac{b x \tanh ^{-1}(c x)}{e}+\frac{d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2}{1+c x}\right )}{e^2}-\frac{d \left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e^2}+\frac{b \log \left (1-c^2 x^2\right )}{2 c e}-\frac{b d \text{Li}_2\left (1-\frac{2}{1+c x}\right )}{2 e^2}+\frac{b d \text{Li}_2\left (1-\frac{2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e^2}\\ \end{align*}

Mathematica [C]  time = 2.44204, size = 315, normalized size = 2.02 \[ \frac{\frac{b \left (c d \text{PolyLog}\left (2,e^{-2 \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-c d \text{PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )+e \sqrt{1-\frac{c^2 d^2}{e^2}} \tanh ^{-1}(c x)^2 e^{-\tanh ^{-1}\left (\frac{c d}{e}\right )}+\frac{1}{2} i \pi c d \log \left (1-c^2 x^2\right )+e \log \left (1-c^2 x^2\right )-2 c d \tanh ^{-1}(c x) \tanh ^{-1}\left (\frac{c d}{e}\right )-2 c d \tanh ^{-1}(c x) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-2 c d \tanh ^{-1}\left (\frac{c d}{e}\right ) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )+2 c d \tanh ^{-1}\left (\frac{c d}{e}\right ) \log \left (i \sinh \left (\tanh ^{-1}\left (\frac{c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )+c d \tanh ^{-1}(c x)^2-i \pi c d \tanh ^{-1}(c x)+2 c d \tanh ^{-1}(c x) \log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )+i \pi c d \log \left (e^{2 \tanh ^{-1}(c x)}+1\right )-e \tanh ^{-1}(c x)^2+2 c e x \tanh ^{-1}(c x)\right )}{c}-2 a d \log (d+e x)+2 a e x}{2 e^2} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(x*(a + b*ArcTanh[c*x]))/(d + e*x),x]

[Out]

(2*a*e*x - 2*a*d*Log[d + e*x] + (b*((-I)*c*d*Pi*ArcTanh[c*x] + 2*c*e*x*ArcTanh[c*x] - 2*c*d*ArcTanh[(c*d)/e]*A
rcTanh[c*x] + c*d*ArcTanh[c*x]^2 - e*ArcTanh[c*x]^2 + (Sqrt[1 - (c^2*d^2)/e^2]*e*ArcTanh[c*x]^2)/E^ArcTanh[(c*
d)/e] + 2*c*d*ArcTanh[c*x]*Log[1 + E^(-2*ArcTanh[c*x])] + I*c*d*Pi*Log[1 + E^(2*ArcTanh[c*x])] - 2*c*d*ArcTanh
[(c*d)/e]*Log[1 - E^(-2*(ArcTanh[(c*d)/e] + ArcTanh[c*x]))] - 2*c*d*ArcTanh[c*x]*Log[1 - E^(-2*(ArcTanh[(c*d)/
e] + ArcTanh[c*x]))] + e*Log[1 - c^2*x^2] + (I/2)*c*d*Pi*Log[1 - c^2*x^2] + 2*c*d*ArcTanh[(c*d)/e]*Log[I*Sinh[
ArcTanh[(c*d)/e] + ArcTanh[c*x]]] - c*d*PolyLog[2, -E^(-2*ArcTanh[c*x])] + c*d*PolyLog[2, E^(-2*(ArcTanh[(c*d)
/e] + ArcTanh[c*x]))]))/c)/(2*e^2)

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Maple [A]  time = 0.112, size = 217, normalized size = 1.4 \begin{align*}{\frac{ax}{e}}-{\frac{ad\ln \left ( cxe+cd \right ) }{{e}^{2}}}+{\frac{bx{\it Artanh} \left ( cx \right ) }{e}}-{\frac{b{\it Artanh} \left ( cx \right ) d\ln \left ( cxe+cd \right ) }{{e}^{2}}}+{\frac{bd\ln \left ( cxe+cd \right ) }{2\,{e}^{2}}\ln \left ({\frac{cxe+e}{-cd+e}} \right ) }+{\frac{bd}{2\,{e}^{2}}{\it dilog} \left ({\frac{cxe+e}{-cd+e}} \right ) }-{\frac{bd\ln \left ( cxe+cd \right ) }{2\,{e}^{2}}\ln \left ({\frac{cxe-e}{-cd-e}} \right ) }-{\frac{bd}{2\,{e}^{2}}{\it dilog} \left ({\frac{cxe-e}{-cd-e}} \right ) }+{\frac{b\ln \left ({c}^{2}{d}^{2}-2\, \left ( cxe+cd \right ) cd+ \left ( cxe+cd \right ) ^{2}-{e}^{2} \right ) }{2\,ce}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a+b*arctanh(c*x))/(e*x+d),x)

[Out]

a/e*x-a*d/e^2*ln(c*e*x+c*d)+b*x*arctanh(c*x)/e-b*arctanh(c*x)*d/e^2*ln(c*e*x+c*d)+1/2*b/e^2*d*ln(c*e*x+c*d)*ln
((c*e*x+e)/(-c*d+e))+1/2*b/e^2*d*dilog((c*e*x+e)/(-c*d+e))-1/2*b/e^2*d*ln(c*e*x+c*d)*ln((c*e*x-e)/(-c*d-e))-1/
2*b/e^2*d*dilog((c*e*x-e)/(-c*d-e))+1/2/c*b/e*ln(c^2*d^2-2*(c*e*x+c*d)*c*d+(c*e*x+c*d)^2-e^2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a{\left (\frac{x}{e} - \frac{d \log \left (e x + d\right )}{e^{2}}\right )} + \frac{1}{2} \, b \int \frac{x{\left (\log \left (c x + 1\right ) - \log \left (-c x + 1\right )\right )}}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x))/(e*x+d),x, algorithm="maxima")

[Out]

a*(x/e - d*log(e*x + d)/e^2) + 1/2*b*integrate(x*(log(c*x + 1) - log(-c*x + 1))/(e*x + d), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b x \operatorname{artanh}\left (c x\right ) + a x}{e x + d}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x))/(e*x+d),x, algorithm="fricas")

[Out]

integral((b*x*arctanh(c*x) + a*x)/(e*x + d), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x \left (a + b \operatorname{atanh}{\left (c x \right )}\right )}{d + e x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*atanh(c*x))/(e*x+d),x)

[Out]

Integral(x*(a + b*atanh(c*x))/(d + e*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{artanh}\left (c x\right ) + a\right )} x}{e x + d}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*arctanh(c*x))/(e*x+d),x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)*x/(e*x + d), x)